a) \(5^{x-1}+5^{x-3}=650\)

\(\Rightarrow5^x\left(\frac{1}{5}+\frac{1}{125}\right)=650\)

\(\Rightarrow5^x=650:\frac{26}{125}\)

\(\Rightarrow5^x=3125\)

\(\Rightarrow5^x=5^5\)

\(\Rightarrow x=5\)

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$

$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$

$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$

$\Leftrightarrow 13x+15=2$

$\Leftrightarrow 13x=2-15=-13$

$\Leftrightarrow x=-13:13=-1$

Bài 2:

$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:

$(y+4)y=5$

$\Leftrightarrow y^2+4y-5=0$

$\Leftrightarrow (y-1)(y+5)=0$

$\Leftrightarrow y=1$ hoặc $y=-5$

Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$

Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$

a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

TH1: \(x\ge5\)

<=> \(\left\{{}\begin{matrix}\left|x-5\right|=x-5\\\left|2x-1\right|=2x-1\end{matrix}\right.\)

PT <=> \(x-5+2x-1=2x+3\)

<=> x = 9 (Tm)

TH2: \(\dfrac{1}{2}\le x< 5\)

<=> \(\left\{{}\begin{matrix}\left|x-5\right|=5-x\\\left|2x-1\right|=2x-1\end{matrix}\right.\)

PT <=> 5 - x + 2x -1 = 2x + 3

<=> x = 1(Tm)

TH3: \(x< \dfrac{1}{2}\)

<=> \(\left\{{}\begin{matrix}\left|x-5\right|=5-x\\\left|2x-1\right|=1-2x\end{matrix}\right.\)

PT <=> \(5-x+1-2x=2x+3\)

<=> \(5x=3< =>x=\dfrac{3}{5}\left(l\right)\)

KL: x \(\in\left\{1;9\right\}\)

c) l x - 5 l = 2x

\(\Leftrightarrow\orbr{\begin{cases}x-5=2x\\x-5=-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2x=5\\x+2x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=5\\3x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)

Hok tốt!!!!!!!

Tìm x, biết:

a) |2x + 1| = 17

<=>\(\orbr{\begin{cases}2x+1=17\\2x+1=-17\end{cases}}\) 

<=>\(\orbr{\begin{cases}2x=16\\2x=-18\end{cases}}\)

<=> \(\hept{\begin{cases}x=8\\x=-9\end{cases}}\)

Phần d không biết , thông cảm

a) Ta có : 

| 2x + 1 | = 17

\(\Rightarrow\orbr{\begin{cases}2x+1=17\\2x+1=-17\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=16\\2x=-18\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)

b) | 3x - 17 | = | 2x - 3 |

Xét các trường hợp :

+) 3x - 17 = - 2x + 3

=> 3x + 2x = 3 + 17

=> 5x = 20

=> x = 4

+) 3x - 17 = 2x - 3

=> 3x - 2x = -3 + 17

=> x = 14

+) -3x + 17 = 2x - 3

=> 2x + 3x = 17 + 3x

=> 5x = 20

=> x = 4

c) | x - 5 | = 2x

\(\Rightarrow\orbr{\begin{cases}x-5=2x\\x-5=-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=5\\3x=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{3}\end{cases}}\)